58th Ukrainian National Mathematical Olympiad

Let's split April into 5 segments. Let from 1 until 5 April he read in average $x$ pages, from 6 until 10 April – in average $a$ pages, from 11 until 20 April – in average $c$ pages, from 21 until 25 April – in average $b$ pages, from 26 until 30 – in average $y$ pages. Then equalities are true: $$\begin{array}{c}\frac{5x+5a+10c}{20}=20,\frac{5a+10c+5b}{20}=30,\frac{10c+5b+5y}{20}=40,\text{ are}\\ x+a+2c=80,a+2c+b=120,2c+b+y=160.\end{array}$$ $$\begin{array}{rl}S& =5x+5a+10c+5b+5y=5(x+a+2c+b+y)=\\ & =5((x+a+2c)+(2c+b+y)-2c)=5(240-2c).\end{array}$$ So, the book has the maximum amount of pages when value $2c$ is the least, and minimal amount of pages, when $2c$ is the biggest.

Case of maximum $S$, or minimal $2c$. As $c\ge 0$, let's try to find possible values of variables for $c=0$. So we have conditions: $$x+a=80,a+b=120,b+y=160.$$ One of possible cases, when $a=0$, then $x=80$, $b=120$ and $y=40$. Then maximum amount of pages equals to $S_{\max }=1200$.

Case of minimal $S$, or maximum $2c$. As $x+a+2c=80$, then $2c\le 80$. Let's try to find possible values of variables for $c=40$. Then we have conditions: $$x+a=0,a+b=40,b+y=80.$$ One of possible cases, when $a=x=0$, then $b=y=40$. Then minimal amount of pages equals to $S_{\min }=800$.